每日一题-week03

day01

将矩阵按对角线排序

快乐模拟

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/*
    将矩阵按对角线排序
    2022-09-12 12:01:07
    by ergou
*/
#include<bits/stdc++.h>
using namespace std;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    int a[105];
    int cnt=0;
    int n,m;
public:
    void so(int x,int y,vector<vector<int>>& mat)
    {
        for(int xx=x,yy=y;xx<n&&yy<m;xx++,yy++,cnt++)
            a[cnt]=mat[xx][yy];

        sort(a,a+cnt);

        for(int xx=x,yy=y,t=0;t<cnt;xx++,yy++,t++)
            mat[xx][yy]=a[t];
    }
    vector<vector<int>> diagonalSort(vector<vector<int>>& mat) {
        n=mat.size();
        m=mat[0].size();
        for(int i=0;i<m;i++){
            cnt=0;
            so(0,i,mat);
        }

        for(int i=1;i<n;i++)
        {
            cnt=0;
            so(i,0,mat);
        }
        return mat;
    }
};
//leetcode submit region end(Prohibit modification and deletion)


signed  main()
{
    Solution s;
    return 0;
}

day02

逃离迷宫

简单搜索

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/*
    2022/9/12 12:31
    by ergou
*/
#include<bits/stdc++.h>

#define ll long long
#define bug(x) cout<<#x<<"x="<<x<<endl
using namespace std;

const int maxx = 1e5 + 5;
const int mod = 1e9 + 7;

inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - 48;
        ch = getchar();
    }
    return x * f;
}
int m,n;
bool mp[105][105];
bool vis[105][105];
int ans[105][105];
int st[4][2]={{1,0},{-1,0},
             {0,1},{0,-1}};
int stx,sty,enx,eny;
int k;
struct no{
    int x,y,cnt,di;
    //0->下, 1->上,2->右,3->左,4->任意
};
void so()
{
    memset(vis,0,sizeof(vis));

    cin>>n>>m;
    char c;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            cin>>c;
            if(c=='.') mp[i][j]=1;
            else mp[i][j]=0;
        }
    //不小心把xy写反了,把这里换一下凑活着用吧
    cin>>k>>sty>>stx>>eny>>enx;


    vis[stx][sty]=1;
    queue<no> q;
    q.push({stx,sty,0,4});
    while(!q.empty())
    {
       no a=q.front();
       if(a.x==enx&&a.y==eny)
       {
           if(a.cnt>k) continue;
           else
           {
               cout<<"yes\n";
               return ;
           }
       }
       q.pop();
       int xx,yy,cnt;
       for(int i=0;i<4;i++)
       {
           xx=a.x+st[i][0];
           yy=a.y+st[i][1];
           if(a.di==4) cnt=0;
           else if(a.di!=i) cnt=a.cnt+1;
           else if(a.di==i) cnt=a.cnt;


           if(cnt>k) continue;
           if(xx>0&&xx<=n&&yy>0&&yy<=m&&mp[xx][yy])
           {
                if(!vis[xx][yy])
                {
                    vis[xx][yy]=1;
                    ans[xx][yy]=cnt;
                    q.push({xx,yy,cnt,i});
                }
                else
                {
                    if(ans[xx][yy]>=cnt)
                    {
                        ans[xx][yy]=cnt;
                        q.push({xx,yy,cnt,i});
                    }
                }
           }
       }
    }

    cout<<"no\n";
    return ;
}

signed main() {
    int t;
    cin>>t;
    while(t--)
    {
        so();
    }
    return 0;
}

day03

将数字变成 0 的操作次数

无聊模拟

这题是怎么混进去的

布吉岛

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/*
    将数字变成 0 的操作次数
    2022-09-12 13:20:22
    by ergou
*/
#include<bits/stdc++.h>
using namespace std;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    int numberOfSteps(int num) {
        int cnt=0;
        while(num)
        {
            if(num&1) num--;
            else num >>= 1;
            cnt++;
        }
        return cnt;
    }
};
//leetcode submit region end(Prohibit modification and deletion)


signed  main()
{
    Solution s;
    return 0;
}

day04

大小为 K 且平均值大于等于阈值的子数组数目

带点双指针的感觉

一开始用的long long,发现内存很丑,试了试int,没爆

好耶

image-20220912135604759 
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/*
    大小为 K 且平均值大于等于阈值的子数组数目
    2022-09-12 13:26:58
    by ergou
*/
#include<bits/stdc++.h>
using namespace std;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    int numOfSubarrays(vector<int>& arr, int k, int threshold) {

        int n=arr.size();
        if(n<k) return 0;
        int num=k*threshold;
        int sum=0;
        int ans=0;
        for(int i=0;i<k;i++) sum+=arr[i];
        if(sum>=num) ans++;

        int l=0,r=k-1;
        while(r+1<n)
        {
            sum-=arr[l++];
            sum+=arr[++r];
            if(sum>=num) ans++;
        }
        return ans;
    }
};
//leetcode submit region end(Prohibit modification and deletion)


signed  main()
{
    Solution s;
    return 0;
}

day05

时钟指针的夹角

快乐算数题

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/*
    时钟指针的夹角
    2022-09-12 13:57:30
    by ergou
*/
#include<bits/stdc++.h>
using namespace std;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public:
    double angleClock(int hour, int minutes) {
        double x=abs((hour*60+minutes)*0.5-minutes*6.0);
        if(x>180) return 360-x;
        return x;
    }
};
//leetcode submit region end(Prohibit modification and deletion)


signed  main()
{
    Solution s;
    return 0;
}

day06

跳跃游戏 IV

一开始以为是记忆化搜索或者剪枝

模了几次觉得不太对劲的样子

看题解居然是bfs…

大意了

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/*
    跳跃游戏 IV
    2022-09-08 17:42:24
    by ergou
*/
#include<bits/stdc++.h>
using namespace std;

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    unordered_map<int,vector<int> > mp;
    bool vis[50005]={0};
    int dis[50005]={0};
    int n;
public:
    int minJumps(vector<int>& arr) {
        n=arr.size();

        for(int i=0;i<n;i++)
        {
            if(mp.count(arr[i])) mp[arr[i]].push_back(i);
            else
            {
                vector<int> a;
                a.push_back(i);
                mp[arr[i]]=a;
            }
        }

        queue<int> q;
        vis[0]=1;
        dis[0]=0;
        q.push(0);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            if(mp.count(arr[u]))
            {
                for(auto i:mp[arr[u]])
                {
                    if(!vis[i])
                    {
                        vis[i]=1;
                        dis[i]=dis[u]+1;
                        q.push(i);
                    }

                }
                //这里要清除,不然如[1,1,...,1,2],这样的样会超时
                //因为压入q中还是会被遍历map,相当于n^2复杂度
                mp.erase(arr[u]);
            }

            if(u+1<n&&!vis[u+1])
            {
                vis[u+1]=1;
                dis[u+1]=dis[u]+1;
                q.push(u+1);
            }
            if(u-1>=0&&!vis[u-1])
            {
                vis[u-1]=1;
                dis[u-1]=dis[u]+1;
                q.push(u-1);
            }
        }
        return dis[n-1];
    }

};
//leetcode submit region end(Prohibit modification and deletion)


signed  main()
{
    Solution s;
    
    return 0;
}